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I am a bit confused about Einstein summation. What should the following expression expand into? (To simplify things, let we have only 2 dimensions)

${a_{00}^i}{b_{i1}^1}$

  1. ${a_{00}^0}{b_{01}^1}+ {a_{00}^1}{b_{11}^1}$ [here we assume the same i for both of the terms].

  2. ${a_{00}^0}{b_{01}^1}+ {a_{00}^0}{b_{11}^1}+ {a_{00}^1}{b_{01}^1}+ {a_{00}^1}{b_{11}^1}$ [All possible combinations].

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  • $\begingroup$ en.wikipedia.org/wiki/Einstein_notation $\endgroup$
    – Ghoster
    Commented Nov 29, 2022 at 18:28
  • $\begingroup$ Think about $\vec a\cdot\vec b=a_ib_i$ for 3D Cartesian vectors. $\endgroup$
    – Ghoster
    Commented Nov 29, 2022 at 18:31
  • $\begingroup$ The important thing to understand is why $a_1b_1+a_2b_2+a_3b_3$ gives a quantity that is invariant under rotations, but $a_1b_1+a_1b_2+\dots+a_3b_2+a_3b_3$ does not. A similar argument applies to Lorentz transformations. Tensor contractions are defined in a way that ensures they produce other tensors. $\endgroup$
    – Ghoster
    Commented Nov 29, 2022 at 20:13

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If we wanted all possible combinations, it would be a bit silly to use the same letter for both indices.

According to the Einstein summation convention, repeated indices are summed over. That is, $$a^i_{\ \ 00} b^1_{\ \ i1} \equiv \sum_i a^i_{\ \ 00} b^1_{\ \ i1} $$ so your first expression is correct.

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